Analytic function $f$ either constant or surjective on unit disk.
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Suppose $f$ is analytic on unit disk and continuous on it. Assume $|f(z)|=1$ on ${x;|x|=1}$. Then $f$ is constant or $f(U)=U$ where $U={x;|x|<1}$.
My attempt:
By maximum principle, $|f(z)|leq 1$ on $U$. Suppose $f(U)neq U$. Then there exists $z_0in U$ such that $z_0notin f(U)$. Then
$$
frac{1}{f(z)-z_0}
$$
is analytic on unit disk. By using maximum principle, then
$$
left|frac{1}{f(z)-z_0}right|leq frac{1}{1-z_0}
$$
on $U$. Thus,
$$
|f(z)-z_0|geq 1-|z_0|
$$
which implies that
$$
f(D)in Usetminus {z;|z-z_0|< 1-z_0}
$$
We can take $z_1in {z;|z-z_0|< 1-z_0}$ such that
$$
|z_1|<|z_0|.
$$
By same argument, on $U$,
$$
|f(z)-z_1|geq 1-|z_1|.
$$
Hence, we can find $|z_n|rightarrow 0$ such that on $U$
$$
|f(z)-z_n|geq 1-|z_n|.
$$
Then we have
$$
|f(z)|geq 1-2|z_n|rightarrow 1.
$$
Hence, $|f(z)|=1$ on $U$. $f$ is constant by using argument of open mapping theorem.
Is this approach correct? I am not sure about $z_n$.
complex-analysis
asked Dec 9 '18 at 16:24
whereamIwhereamI
322115
$endgroup$
|
show 1 more comment
$begingroup$
Suppose $f$ is analytic on unit disk and continuous on it. Assume $|f(z)|=1$ on ${x;|x|=1}$. Then $f$ is constant or $f(U)=U$ where $U={x;|x|<1}$.
My attempt:
By maximum principle, $|f(z)|leq 1$ on $U$. Suppose $f(U)neq U$. Then there exists $z_0in U$ such that $z_0notin f(U)$. Then
$$
frac{1}{f(z)-z_0}
$$
is analytic on unit disk. By using maximum principle, then
$$
left|frac{1}{f(z)-z_0}right|leq frac{1}{1-z_0}
$$
on $U$. Thus,
$$
|f(z)-z_0|geq 1-|z_0|
$$
which implies that
$$
f(D)in Usetminus {z;|z-z_0|< 1-z_0}
$$
We can take $z_1in {z;|z-z_0|< 1-z_0}$ such that
$$
|z_1|<|z_0|.
$$
By same argument, on $U$,
$$
|f(z)-z_1|geq 1-|z_1|.
$$
Hence, we can find $|z_n|rightarrow 0$ such that on $U$
$$
|f(z)-z_n|geq 1-|z_n|.
$$
Then we have
$$
|f(z)|geq 1-2|z_n|rightarrow 1.
$$
Hence, $|f(z)|=1$ on $U$. $f$ is constant by using argument of open mapping theorem.
Is this approach correct? I am not sure about $z_n$.
complex-analysis
asked Dec 9 '18 at 16:24
whereamIwhereamI
322115
$endgroup$
$begingroup$
Although you have constructed a sequence $z_n$ that has strictly decreasing magnitude, as written you have not shown in converges to zero specifically.
$endgroup$
– eyeballfrog
Dec 9 '18 at 17:04
$begingroup$
@eyeballfrog if i take $z_1=z_0-frac{1-|z_0|}{2}frac{z_0}{|z_0|}$, then $z_n$ should converges to $0$. Is that enough?
$endgroup$
– whereamI
Dec 9 '18 at 17:06
$begingroup$
See math.stackexchange.com/questions/116965/… for some solutions.
$endgroup$
– Martin R
Dec 9 '18 at 17:07
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@MartinR thanks a lot
$endgroup$
– whereamI
Dec 9 '18 at 17:08
$begingroup$
@whereamI Yes. In fact, it reaches zero in a finite number of steps, since once $|z_n| < 1/2$ you can choose $z_{n+1} = 0$.
$endgroup$
– eyeballfrog
Dec 9 '18 at 17:13
|
show 1 more comment
$begingroup$
Suppose $f$ is analytic on unit disk and continuous on it. Assume $|f(z)|=1$ on ${x;|x|=1}$. Then $f$ is constant or $f(U)=U$ where $U={x;|x|<1}$.
My attempt:
By maximum principle, $|f(z)|leq 1$ on $U$. Suppose $f(U)neq U$. Then there exists $z_0in U$ such that $z_0notin f(U)$. Then
$$
frac{1}{f(z)-z_0}
$$
is analytic on unit disk. By using maximum principle, then
$$
left|frac{1}{f(z)-z_0}right|leq frac{1}{1-z_0}
$$
on $U$. Thus,
$$
|f(z)-z_0|geq 1-|z_0|
$$
which implies that
$$
f(D)in Usetminus {z;|z-z_0|< 1-z_0}
$$
We can take $z_1in {z;|z-z_0|< 1-z_0}$ such that
$$
|z_1|<|z_0|.
$$
By same argument, on $U$,
$$
|f(z)-z_1|geq 1-|z_1|.
$$
Hence, we can find $|z_n|rightarrow 0$ such that on $U$
$$
|f(z)-z_n|geq 1-|z_n|.
$$
Then we have
$$
|f(z)|geq 1-2|z_n|rightarrow 1.
$$
Hence, $|f(z)|=1$ on $U$. $f$ is constant by using argument of open mapping theorem.
Is this approach correct? I am not sure about $z_n$.
complex-analysis
asked Dec 9 '18 at 16:24
whereamIwhereamI
322115
$endgroup$
Suppose $f$ is analytic on unit disk and continuous on it. Assume $|f(z)|=1$ on ${x;|x|=1}$. Then $f$ is constant or $f(U)=U$ where $U={x;|x|<1}$.
My attempt:
By maximum principle, $|f(z)|leq 1$ on $U$. Suppose $f(U)neq U$. Then there exists $z_0in U$ such that $z_0notin f(U)$. Then
$$
frac{1}{f(z)-z_0}
$$
is analytic on unit disk. By using maximum principle, then
$$
left|frac{1}{f(z)-z_0}right|leq frac{1}{1-z_0}
$$
on $U$. Thus,
$$
|f(z)-z_0|geq 1-|z_0|
$$
which implies that
$$
f(D)in Usetminus {z;|z-z_0|< 1-z_0}
$$
We can take $z_1in {z;|z-z_0|< 1-z_0}$ such that
$$
|z_1|<|z_0|.
$$
By same argument, on $U$,
$$
|f(z)-z_1|geq 1-|z_1|.
$$
Hence, we can find $|z_n|rightarrow 0$ such that on $U$
$$
|f(z)-z_n|geq 1-|z_n|.
$$
Then we have
$$
|f(z)|geq 1-2|z_n|rightarrow 1.
$$
Hence, $|f(z)|=1$ on $U$. $f$ is constant by using argument of open mapping theorem.
Is this approach correct? I am not sure about $z_n$.
complex-analysis
complex-analysis
asked Dec 9 '18 at 16:24
whereamIwhereamI
322115
asked Dec 9 '18 at 16:24
whereamIwhereamI
322115
asked Dec 9 '18 at 16:24
whereamIwhereamI
322115
asked Dec 9 '18 at 16:24
whereamIwhereamI
322115
asked Dec 9 '18 at 16:24
whereamIwhereamI
322115
322115
$begingroup$
Although you have constructed a sequence $z_n$ that has strictly decreasing magnitude, as written you have not shown in converges to zero specifically.
$endgroup$
– eyeballfrog
Dec 9 '18 at 17:04
$begingroup$
@eyeballfrog if i take $z_1=z_0-frac{1-|z_0|}{2}frac{z_0}{|z_0|}$, then $z_n$ should converges to $0$. Is that enough?
$endgroup$
– whereamI
Dec 9 '18 at 17:06
$begingroup$
See math.stackexchange.com/questions/116965/… for some solutions.
$endgroup$
– Martin R
Dec 9 '18 at 17:07
$begingroup$
@MartinR thanks a lot
$endgroup$
– whereamI
Dec 9 '18 at 17:08
$begingroup$
@whereamI Yes. In fact, it reaches zero in a finite number of steps, since once $|z_n| < 1/2$ you can choose $z_{n+1} = 0$.
$endgroup$
– eyeballfrog
Dec 9 '18 at 17:13
|
show 1 more comment
$begingroup$
Although you have constructed a sequence $z_n$ that has strictly decreasing magnitude, as written you have not shown in converges to zero specifically.
$endgroup$
– eyeballfrog
Dec 9 '18 at 17:04
$begingroup$
@eyeballfrog if i take $z_1=z_0-frac{1-|z_0|}{2}frac{z_0}{|z_0|}$, then $z_n$ should converges to $0$. Is that enough?
$endgroup$
– whereamI
Dec 9 '18 at 17:06
$begingroup$
See math.stackexchange.com/questions/116965/… for some solutions.
$endgroup$
– Martin R
Dec 9 '18 at 17:07
$begingroup$
@MartinR thanks a lot
$endgroup$
– whereamI
Dec 9 '18 at 17:08
$begingroup$
@whereamI Yes. In fact, it reaches zero in a finite number of steps, since once $|z_n| < 1/2$ you can choose $z_{n+1} = 0$.
$endgroup$
– eyeballfrog
Dec 9 '18 at 17:13
$begingroup$
Although you have constructed a sequence $z_n$ that has strictly decreasing magnitude, as written you have not shown in converges to zero specifically.
$endgroup$
– eyeballfrog
Dec 9 '18 at 17:04
$begingroup$
Although you have constructed a sequence $z_n$ that has strictly decreasing magnitude, as written you have not shown in converges to zero specifically.
$endgroup$
– eyeballfrog
Dec 9 '18 at 17:04
$begingroup$
@eyeballfrog if i take $z_1=z_0-frac{1-|z_0|}{2}frac{z_0}{|z_0|}$, then $z_n$ should converges to $0$. Is that enough?
$endgroup$
– whereamI
Dec 9 '18 at 17:06
$begingroup$
@eyeballfrog if i take $z_1=z_0-frac{1-|z_0|}{2}frac{z_0}{|z_0|}$, then $z_n$ should converges to $0$. Is that enough?
$endgroup$
– whereamI
Dec 9 '18 at 17:06
$begingroup$
See math.stackexchange.com/questions/116965/… for some solutions.
$endgroup$
– Martin R
Dec 9 '18 at 17:07
$begingroup$
See math.stackexchange.com/questions/116965/… for some solutions.
$endgroup$
– Martin R
Dec 9 '18 at 17:07
$begingroup$
@MartinR thanks a lot
$endgroup$
– whereamI
Dec 9 '18 at 17:08
$begingroup$
@MartinR thanks a lot
$endgroup$
– whereamI
Dec 9 '18 at 17:08
$begingroup$
@whereamI Yes. In fact, it reaches zero in a finite number of steps, since once $|z_n| < 1/2$ you can choose $z_{n+1} = 0$.
$endgroup$
– eyeballfrog
Dec 9 '18 at 17:13
$begingroup$
@whereamI Yes. In fact, it reaches zero in a finite number of steps, since once $|z_n| < 1/2$ you can choose $z_{n+1} = 0$.
$endgroup$
– eyeballfrog
Dec 9 '18 at 17:13
|
show 1 more comment
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$begingroup$
Although you have constructed a sequence $z_n$ that has strictly decreasing magnitude, as written you have not shown in converges to zero specifically.
$endgroup$
– eyeballfrog
Dec 9 '18 at 17:04
$begingroup$
@eyeballfrog if i take $z_1=z_0-frac{1-|z_0|}{2}frac{z_0}{|z_0|}$, then $z_n$ should converges to $0$. Is that enough?
$endgroup$
– whereamI
Dec 9 '18 at 17:06
$begingroup$
See math.stackexchange.com/questions/116965/… for some solutions.
$endgroup$
– Martin R
Dec 9 '18 at 17:07
$begingroup$
@MartinR thanks a lot
$endgroup$
– whereamI
Dec 9 '18 at 17:08
$begingroup$
@whereamI Yes. In fact, it reaches zero in a finite number of steps, since once $|z_n| < 1/2$ you can choose $z_{n+1} = 0$.
$endgroup$
– eyeballfrog
Dec 9 '18 at 17:13